PDA

View Full Version : Apologies and basic question



mallkins
December 21st 2005, 10:40 PM
Firstly, apologies. It has been many years since I last took a probabilities course, and so I know this will be a hard slog through to Feb exam.

Now to basic question:

In the example where it talks of rolling a die there are 64 subsets, or 2^6. It is obvious where the 6 comes from, but where is the 2 from?

Ken
December 21st 2005, 11:22 PM
You talk of this example like we all know about. What are you talking about?

mallkins
December 21st 2005, 11:34 PM
Hi Ken

Sorry, the way it is presented in the book is as if it is a standard example (as you may see from another thread, the book seems somewhat poor at explaining things).

The example reads

A standard die with six distinguishable sides is rolled and the outcome is observed. The sample space for this experiment is

S = {1,2,3,4,5,6}

The event space for this experiment is

E = {0,{1} etc.

and it goes on to list all 64 subsets.

All this is fine but the next line says.

Note that S has 2^6 = 64 subsets! (i.e. 2 to the power of 6)

The 6 is obvious, but why the 2 to the power of 6? Where does the 2 come from?

A more relevant question may be do we need to cover event spaces?

Thanks,

Miles


PS. Does anybody know how to do superscript in these discussion windows?

Ken
December 22nd 2005, 12:13 AM
Still not exactly sure what's going on here, but I'm assuming it's something like you roll a die and then flip a coin. There's then 64 outcomes.

krzysio
December 22nd 2005, 12:29 AM
The example reads
A standard die with six distinguishable sides is rolled and the outcome is observed. The sample space for this experiment is
S = {1,2,3,4,5,6}
The event space for this experiment is
E = {0,{1} etc.
and it goes on to list all 64 subsets.
All this is fine but the next line says.
Note that S has 2^6 = 64 subsets! (i.e. 2 to the power of 6)
The 6 is obvious, but why the 2 to the power of 6? Where does the 2 come from?
A more relevant question may be do we need to cover event spaces?
Thanks,
Miles


A useful and important thing to remember: for any set of n elements, the number of its subsets of 2^n (2 to the power n). To prove this, note that there are n choose 0 subsets with 0 elements, n choose 1 subsets with 1 element, ..., n choose n subsets with n elements. And the total of those is:
(n choose 0)*1^0*1^n + (n choose 1)*1^1*1^(n-1)+....+ (n choose n)*1^n*1^0 = (1 + 1)^n = 2^n. QED.

And, of course, this is covered in my ASM Manual for exam P, available from actexmadriver.com.

Good luck.
Yours,
Krzys'